一道高中三角函数的计算2sin50°+sin80°（1+根号（1+根号三tan10°））除以根号（1+cos10°）虽然看着很费劲,但无奈打不出根号,大家就此将就点,谢谢了.

你的题目应该是
2sin50°+sin80°(1+√3*tan10°)/√(1+cos10°)
2sin50°+sin80°(1+√3*tan10°)/√(1+cos10°)
={2sin50°+cos10°[1+(cos30°/sin30°)(sin10°/cos10°)]}/ √(2cos^2 5°)
={2sin50°+cos10°[1+(cos30°sin10°)/(sin30°cos10°)]}/ (√2*cos5°)
={2sin50°+cos10°[(cos30°sin10°+sin30°cos10°)/(sin30°cos10°)]}/ (√2*cos5°)
={2sin50°+cos10°[sin40°/(sin30°cos10°)]}/ (√2*cos5°)
=[2sin50°+(sin40°/sin30°)]/ (√2*cos5°)
=[2sin50°+2sin40°]/ (√2*cos5°)
=2√2*sin(40°+45°)/ (√2*cos5°)
=2sin85°/cos5°
=2